Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<iostream>
#include<iomanip>
#include<math.h>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<random>
#include<time.h>
#define INF 1000000000ll
#define MOD 1000000007ll
#define EPS 1e-10
#define REP(i,m) for(long long i=0; i<(ll)m; i++)
#define FOR(i,n,m) for(long long i=n; i<(ll)m; i++)
#define DUMP(a) for(long long dump=0; dump<(ll)a.size(); dump++) { cout<<a[dump]; if(dump!=(ll)a.size()-1) cout<<" "; else cout<<endl; }
#define ALL(v) v.begin(),v.end()
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef long double ld;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n,m,s,t;
	cin>>n>>m>>s>>t;
	s--;
	t--;
	vector<vector<ll>> adj(n);
	REP(i,m) {
		ll x,y;
		cin>>x>>y;
		x--;
		y--;
		adj[x].pb(y);
		adj[y].pb(x);
	}
	vector<ll> froms(n,INF);
	vector<ll> fromt(n,INF);
	froms[s]=0;
	fromt[t]=0;
	queue<ll> q;
	q.push(s);
	while(!q.empty()) {
		ll p=q.front();
		q.pop();
		REP(i,adj[p].size()) {
			if(froms[adj[p][i]]>froms[p]+1) {
				froms[adj[p][i]]=froms[p]+1;
				q.push(adj[p][i]);
			}
		}
	}
	q.push(t);
	while(!q.empty()) {
		ll p=q.front();
		q.pop();
		REP(i,adj[p].size()) {
			if(fromt[adj[p][i]]>fromt[p]+1) {
				fromt[adj[p][i]]=fromt[p]+1;
				q.push(adj[p][i]);
			}
		}
	}
	vector<ll> cnt1(n,0);
	vector<ll> cnt2(n,0);
	REP(i,n) if(froms[i]!=INF) cnt1[froms[i]]++;
	REP(i,n) if(fromt[i]!=INF) cnt2[fromt[i]]++;
	ll ans=0;
	REP(i,n) {
		if(froms[t]-i-2>=0) ans+=cnt1[i]*cnt2[froms[t]-i-2];
	}
	cout<<ans<<endl;
}